Problem: $\int \left( \dfrac{- x^{2} + x}{x^4}\right)dx=$ $+C$
The integrand is the quotient of two functions: $- x^{2} + x$ and $x^4$. Although it is tempting to take the quotient of their integrals, this would not work. $\int \dfrac{f(x)}{g(x)}\,dx\neq\dfrac{\int f(x)\,dx}{\int g(x)\,dx}$ Instead, what we should do is divide each term in the numerator by the denominator. $\begin{aligned} &\phantom{=}\int \left( \dfrac{- x^{2} + x}{x^4}\right)dx \\\\ &=\int\left(-\dfrac{x^{2}}{x^4}+\dfrac{x}{x^4}\right)dx \\\\ &=\int (- x^{-2} + x^{-3} )\,dx \end{aligned}$ Now we can integrate using the reverse power rule, the sum rule, and the constant multiple rule for indefinite integrals. $\begin{aligned} &\phantom{=}\int \left( \dfrac{- x^{2} + x}{x^4}\right)dx \\\\ &=\int (- x^{-2} + x^{-3})\,dx \\\\ &= -\int x^{-2}\,dx +\int x^{-3}\,dx \\\\ &=-\dfrac{x^{-1}}{-1} +\dfrac{x^{-2}}{-2} +C \\\\ &= \dfrac{1}{x} -\dfrac{1}{2x^2} +C \end{aligned}$ In conclusion, $\int \left( \dfrac{- x^{2} + x + }{x^4}\right)dx= \dfrac{1}{x} -\dfrac{1}{2x^2} +C$